Bourbaki

Proposition 3

Tag 03HK, permanent. Cite it as [Bourbaki, Tag 03HK].

Theory of Sets, chapter II, § 3. Correspondences

Let G, G′ be two graphs. The inverse of G′ ∘ G is then G−1∘G′−1\overset{-1}{G} \circ \overset{-1}{G'}.

For the relation "(x,y)∈G(x, y) \in G and (y,z)∈G′(y, z) \in G'" is equivalent to

(z,y)∈G′−1 and (y,x)∈G−1.(z, y) \in \overset{-1}{G'} \text{ and } (y, x) \in \overset{-1}{G}.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2004, Springer. Not translated.